From the examination of the expressions of the polynomials, I drew
some conclusions:

(1) Odd order polynomials contain only even powers of $x$,
including zero.

(2) Even order polynomials contain only odd powers of $x$.

(3) There are alternate signs of terms in each polynomial, starting
with the first, of order $(n-1)$ for $P_n(x)$, which is positive.

I have also shown, as required in the question, that $P_4(x)$
contains as a factor $P_2(x)$, $P_6(x)$ contains as factor $P_3(x)$
(that is every root of $P_3$ is a root of $P_6$), $P_8(x)$ contains
as factor $P_4(x)$ and $P_{10}(x)$ contains as factor $P_5(x$.

Editor's note: This suggests a conjecture that
$P_{2k}=P_k(P_{k+1}-P_{k-1})$ where $k$ is any natural number. This
is true but the general proof is beyond the scope of school
mathematics. Andrei made further observations about the
coefficients in these polynomials in the hope of finding explicit
formulae for the $n$th order polynomial. He found, looking at
Fibonacci numbers, that these polynomials are very similar to
Fibonacci polynomials, which are given by the recursive relation:

F_{n+2}(x) = xF_{n+1}(x) + Fn(x) with $F_0(x) = 0$ and $F_1(x) =
1$. Using these relations, he found the first 10 Fibonacci
polynomials, which are, up to the signs found in $P_n(x)$,
identical with $F_n(x)$.

Using the explicit formula of Fibonacci polynomials from
http://mathworld.wolfram.com, Andrei hoped to write correctly the
explicit formula for the polynomials in the problem, as:

$$P_n(x)=\sum_{j=0}^{(n-1)/2}(-1)^j C^{n-j-1}_jx^{n-2j-1}.$$

The formula works well up to $n = 10$. From this formula, all
properties could be found easily, although Andrei was not able to
demonstrate the general case that $P_{2n}(x)$ is divisible by
$P_n(x)$.

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